NEET Physics Current Electricity MCQs Set 25

Answer: (d) 2.5\(℃\)

Explanation:
1. If \(T\) is the smallest temperature difference that can be detected, then
2. \(40×{10}^{-6}T=100×{10}^{-6}\)
3. \(⇒T=2.5℃\)

Answer: (b) Decreases with the increase of temperature

Explanation:
1. Because as temperature increases, the resistivity increases and hence the relaxation time decreases for conductors \(\left(τ∝\frac{1}{ρ}\right)\)

Answer: (b) 30 W

Explanation:
1. \({P}_{1}={P}_{2}=60\) W; when bulbs are connected in series then total power
2. \({P}_{s}=\frac{{P}_{1}{P}_{2}}{{P}_{1}+{P}_{2}}=\frac{60×60}{60+60}=\)30 W

Answer: (a) \(40℃\)

Explanation:
1. Neutral temperature \({T}_{n}=\frac{{T}_{c} + {T}_{f}}{2}\)
2. \(320°=\frac{{T}_{c}+ 600°}{2}\)
3. \(640°={T}_{c}+600°\)
4. \({T}_{c}=40℃\)

Answer: (d) \(68℃\)

Explanation:
1. \({i}^{2}Rt=ms ∆t\)
2. \(⇒ {\left(4\right)}^{2}×7×3×60\)
3. \(=0.1×4.2×{10}^{3}×\left[T-\left(20+273\right)\right]\)
4. \(∴ T=341 K=68℃\)

Answer: (c) Decreases

Explanation:
1. For semiconductor the temperature coefficient of resistance (\(α\)) is negative. Hence, resistivity will decrease with the temperature rise.

Answer: (d) 9 :1

Explanation:
1. \(S=\frac{{i}_{g}G}{(i-{i}_{g})}⇒\frac{G}{S}=\frac{i={i}_{g}}{{i}_{g}}=\frac{10-1}{1}=\frac{9}{1}\)

Answer: (a) \(25 watt\) bulb

Explanation:
1. The resistance of \(25 W\) bulb is greater than \(100 W\) bulb. So for the same current, heat produced will be more in \(25 W\) bulb. So it will glow more brightly

Answer: (b) Yellow, violet, green, gold

Explanation:
1. Yellow, Violet and Gold

Answer: (c) \(9:1\)

Explanation:
1. \(\frac{{J}_{1}}{{J}_{2}}=\frac{I/π{r}_{1}^{2}}{I/π{r}_{2}^{2}}={\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{2}={\left(\frac{3}{1}\right)}^{2}=\frac{9}{1}\)

Answer: (c) \(0.4 mm/s\)

Explanation:
1. \({V}_{d}=\frac{i}{nAe}=\frac{5.4}{8.4×{10}^{28}×{10}^{-6}×1.6×{10}^{-19}}\)
2. \(=0.4×{10}^{-3} m/sec=0.4 mm/sec\)

Answer: (c) 53 A

Explanation:
1. Current, \(i=\frac{dQ}{dt}=\frac{d}{dt}\left(5{t}^{2}+3t+1\right)=10t+3\)
2. When t=5s
3. \(i=10×5+3=53A\)

Answer: (c) \(0.1 Ω\)

Explanation:
1. \({R}^{'}=\frac{R}{n}=\frac{1}{10}=0.1Ω\)

Answer: (d) Potentiometer

Explanation:
1. For greater sensitivity of meter bridge the resistance (R) taken in the resistance box should be such that the null point is nearly in the middle of the wire. In this position all resistance P, Q, R and S become nearly equal. The emf of cell depends upon the size and area of electrodes.

Answer: (d) \(\frac{R}{25}\)

Explanation:
1. \(R∝l\)
2. Hence every new piece will have a resistance \(\frac{R}{10}.\) If two pieces are connected in series, then their resistance \(=\frac{2R}{10}=\frac{R}{5}\)
3. If 5 such combinations are joined in parallel, then net resistance \(=\frac{R}{5×5}=\frac{R}{25}\)

Answer: (c) \(>25 W\)

Explanation:
1. If resistance does not vary with temperature \(P\) consumed
2. \(={\left(\frac{{V}_{A}}{{V}_{R}}\right)}^{2}×{P}_{R}={\left(\frac{110}{220}\right)}^{2}×100=25W\). But in second case resistance decreases so consumed power will be more than \(25 W\)

Answer: (b) 50 W

Explanation:
1. Let \(R\) be the resistance of each bulb.
2. We have \(P=\frac{{V}^{2}}{R}\)
3. or \(R=\frac{{V}^{2}}{P}=\frac{{V}^{2}}{100}\)
4. When the bulbs are connected in series, the voltage across each of them is \(V/2\). Hence, the total power consumed is
5. \({P}^{'}=\frac{{\left(\frac{V}{2}\right)}^{2}}{R}+\frac{{\left(\frac{V}{2}\right)}^{2}}{R}=\frac{{V}^{2}}{2R}\)
6. \({∴P}^{'}=\frac{P}{2}=\frac{100}{2}=\)50 W

Answer: (b) 1.188

Explanation:
1. \(\frac{{E}_{1}}{{E}_{2}}=\frac{{l}_{1}}{{l}_{2}}⟹\frac{1.08}{{E}_{2}}=\frac{400}{440}\)
2. \(⟹{E}_{2}=\frac{440×1.08}{400}=1.188V.\)

Answer: (a) \(C\)

Explanation:
1. Temperature coefficient of a semiconductor is negative

Answer: (a) \(12 S\)

Explanation:
1. \({R}_{1}-\frac{1}{2Ω},{R}_{2}=\frac{1}{4Ω};{R}_{3}=1/6Ω\)
2. In parallel; \(\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{R3 }=2+4+6=12\)
3. or Equivalent conductance, \(G=\frac{1}{{R}_{p}}=12S\)

Answer: (a) 25%

Explanation:
1. Thermo electric power, \(S∝θ\)
2. \(∴\frac{{S}_{100}}{{S}_{80}}=\frac{100}{80}\mathrm{or}{S}_{100}={S}_{80}×\frac{100}{80}=\frac{5}{4}{S}_{80}\)
3. Therefore % change in thermo electric power
4. \(=\left(\frac{{S}_{100}-{S}_{80}}{{S}_{80}}\right)×100\)
5. \(=\left(\frac{\frac{5}{4}{S}_{80}-{S}_{80}}{{S}_{80}}\right)×100\)
6. = 25%

Answer: (c) \(4V\)

Explanation:
1. \({E}_{1}=?, {l}_{1}=60cm;{E}_{2}=3V, {l}_{2}=45cm\)
2. In balance condition
3. \(\frac{{E}_{1}}{{E}_{2}}=\frac{{l}_{1}}{{l}_{2}}⇒\frac{{E}_{1}}{3}=\frac{60}{45}⇒{E}_{1}=4 volt\)

Answer: (b) \(10Ω\)

Explanation:
1. \(r=\left(\frac{{l}_{1}-{l}_{2}}{{l}_{2}}\right)×{R}^{'}=\left(\frac{{l}_{1}-2}{2}\right)×5\) …(i)
2. and \(r=\left(\frac{{l}_{1}-3}{3}\right)×10\) …(ii)
3. on solving (i) and (ii), \(r=10 Ω\)

Answer: (d) Both (b) and (c)

Explanation:
1. In the normal condition current flows from \(X\) to \(Y\) through cold junction. After increasing the temperature of hot junction beyond temperature of inversion the current is reversed \(i.e. X\) to \(Y\) through hot junction or\(Y\) to \(X\) through cold junction

Answer: (c) \(1/3Ω\)

Explanation:
1. \(2=\frac{ε}{2+r}\)
2. \(0.5=\frac{ε}{9+r}\) or\(\frac{2}{0.5}=\frac{9+r}{2+r} ∴r=\frac{1}{3}Ω\)

Answer: (b) The resistance of an incandescent lamp is greater when the lamp is switched off

Explanation:
1. As temperature increases resistance of filament also increase

Answer: (b) Transport current flows through the entire area of cross-section of the wire

Explanation:
1. Supercurrent always flows on the surface of the superconductor.

Answer: (c) \(3×{\left(\frac{3}{2}\right)}^{3/2}\)A

Explanation:
1. For a fuse \({I}^{2}∝{r}^{3}\)
2. \(∴ \frac{{I}_{1}^{2}}{{I}_{2}^{2}}=\frac{{r}_{1}^{3}}{{r}_{2}^{2}}\)
3. \(\frac{{3}^{2}}{{I}_{2}^{2}}={\left(\frac{0.02}{0.03}\right)}^{3}\)
4. \({I}_{2}=3×{\left(\frac{3}{2}\right)}^{3/2}\)A

Answer: (c) 3/2

Explanation:
1. In series,
2. \({P}_{s}=\frac{{P}_{1} ×{P}_{2}}{{P}_{1}+{P}_{2}}=12 W\)
3. In parallel, \({P}_{p}={P}_{1}+P{P}_{2}=50 W\)
4. \(∴ {P}_{1}{P}_{2}=12×50=600\)
5. Now, \({\left({P}_{1}-{P}_{2}\right)}^{2}={\left({P}_{1}+{P}_{2}\right)}^{2}-4{P}_{1}{P}_{2}\)
6. \(={\left(50\right)}^{2}-4×600\)
7. \(=2500-2400=100\)
8. \(∴ {P}_{1}-{P}_{2}=10\)
9. Or \({P}_{1}=30 W, {P}_{2}=20 W\)
10. \(∴ \frac{{P}_{1}}{{P}_{2}}=\frac{3}{2}\)

Answer: (a) \(\frac{{I}^{2}Rt}{4.2}\)

Explanation:
1. \(H=\frac{{I}^{2}Rt}{J}=\frac{{I}^{2}Rt}{4.2}\mathrm{cal}\)

Answer: (d) None is damaged

Explanation:
1. Due to high resistance of voltmeter, connected in series, the effective resistance of circuit will increase and hence the current in circuit will decrease. Due to which the ammeter and voltmeter will not be damaged.

Answer: (d) 6.8 mg

Explanation:
1. \({m}_{\mathrm{Ag}}={m}_{\mathrm{Cu}}×{E}_{\mathrm{Ag}}/{E}_{\mathrm{Cu}}\)
2. \(=2×108/(63.6\)/2) = 6.8 \(mg\).